how much work is required to stretch a spring 0.133 m if its force constant is 9.17N/m?

Question

anonymous · Answer

it will be 0.5kx^2 
where k=9.17&x=0.133

anonymous · Answer

$$W = F * D$$

valentin68 · Answer

By definition the work needed to stretch a sping having elastic constant k with the dispăplacement x is equal to the elastic potential energy variation W =k*x^2/2 = 9.17*0.133*0.133/2 =0.081 J

anonymous · Answer

This problem is in the Energy Conservation in Oscilatory Motion section of the textbook why is that?

anonymous · Answer

this is because the work we do on the spring gets converted to the potential energy of the spring & gets conserved..and can be used when needed:)

anonymous · Answer

.16220j

anonymous · Answer

i agree vth 009infi...

anonymous · Answer

@Decart you got my point???

anonymous · Answer

Yes the book likes to throw you of course work was chapter 7 so I did not expect to see it here.'

anonymous · Answer

but how can it be so 1/2mv^2 is for energy@valentin68